For the following reaction, KClO4—>KCl+2O2, assign oxidation states to each element on each side of the eq.
ClO4= -1 From this you can get…
O= 0 (because it is by itself)
Chlorine is reduced and Oxygen is oxidized (OILRIG- Oxidation is Losing electrons, Reduction is Gaining electrons)
Source(s): AP chem student
K will always be a +1, and oxygen will always be a -2. It is the chlorine that can have a number of different states. And you need a neutral molecule. But with diatomic oxygen, it is neutral, so one really doesn’t think in terms of assigning a number.
As far as which is oxidized and which is reduced, it has to do with which way the oxidation number moves, and you need to commit that idea to memory.
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